leetcode Daily Challenge on November 5th, 2020.
Difficulty : Easy
Related Topics : Array、Math、Greedy
We have
n
chips, where the position of thei
th
chip isposition[i]
.We need to move all the chips to the same position. In one step, we can change the position of the
i
th
chip fromposition[i]
to:
position[i] + 2
orposition[i] - 2
withcost = 0
position[i] + 1
orposition[i] - 1
withcost = 1.
Return the minimum cost needed to move all the chips to the same position.Input: position = [1,2,3] Output: 1 Explanation: First step: Move the chip at position 3 to position 1 with cost = 0. Second step: Move the chip at position 2 to position 1 with cost = 1. Total cost is 1.
Input: position = [2,2,2,3,3] Output: 2 Explanation: We can move the two chips at poistion 3 to position 2. Each move has cost = 1. The total cost = 2.
Input: position = [1,1000000000] Output: 1
1 <= position.length <= 100
1 <= position[i] <= 10^9
- mine
- Java
Runtime: 0 ms, faster than 100.00%, Memory Usage: 36.5 MB, less than 7.61% of Java online submissions
public int minCostToMoveChips(int[] position) { int odd = 0, even = 0; for(int i : position){ if(i % 2 == 0){ even++; }else{ odd++; } } return Math.min(even, odd); }
- Java